Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Determinants

Question:

The value of the integral $I=\int\limits_{-1}^1\left(x+x^3+x^5\right) dx$ is :

Options:

2

-2

0

1

Correct Answer:

0

Explanation:

$I=\int\limits_{-1}^1\left(x+x^3+x^5\right) dx$

for odd functions

$\int\limits_{-a}^{a} f(x) dx = 0$

Let $\int f(x) dx = F(x)$

so if f(x) is odd. F(x) is even

so $\int\limits_{-a}^{a} f(x) dx = F(a) - F(-a)$

= 0

as (F(x) is even so F(x) = F(-x))

⇒  I = 0 + 0 + 0

= 0

since all are off function